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← Session 31 — ii′-5f: tight R-bound 10.5/48 (piece 3 + pinch)

#1819 assistant 2026-08-24 20:08:47.202 UTC · 1196→1186 tok open ⤢
thinking (2189 chars)
REPRODUCED in isolation! The timeout is at 17:27 — `hAB.add (testC.mul_left _)` — so the .add itself times out. The hAB has an explicit ascribed type; hAB.add (testC.mul_left _) — the result type: Summable (fun x => (c0·A x + c1·B x) + c2·C x) vs hABC's ascription — the isDefEq times out. WHY? Let me think about `Summable.add`: `theorem Summable.add {f g : ι → M} (hf : Summable f) (hg : Summable g) : Summable fun x => f x + g x` — result: fun x => f x + g x with f,g as the implicit from the proofs. hAB : Summable (fun ℓ => c0·A ℓ + c1·B ℓ) — the ASCIBED type — so f := fun ℓ => c0·A ℓ + c1·B ℓ. testC.mul_left _ : Summable (fun x => ?c·C x) — with the underscore — the underscore ! The `testC.mul_left _` has a METAVARIABLE ?c — and when unifying hABC's ascription (0.886 coefficient), it should resolve... but maybe the elaborator tries to unify `fun x => ?c * C x` with the expected `fun x => 0.886 * C x` and the isDefEq on... — that should be trivial. OR: the result of hAB.add hC : Summable (fun x => hAB-fun x + hC-fun x) — with hAB-fun = the ascribed lambda and hC-fun = fun x => ?c·C x: the result's body: (c0·A x + c1·B x) + ?c·C x — vs ascription (c0·A x + c1·B x) + 0.886·C x — unify ?c := 0.886 ✓ — the isDefEq of the two big lambdas — the problem might be that hAB's ascribed TYPE is not reducibly equal to the body form... Hmm wait — actually the deeper issue could be `testC.mul_left _`: Summable.mul_left's signature: `Summable.mul_left {f : ι → M} (r : M) (hf : Summable f) : Summable fun x => r * f x`? — or is r implicit/explicit? If `Summable.mul_left (r : M) (hf : Summable f)` — testC.mul_left _ — the FIRST explicit arg is r (a number, the underscore) and hf = testC ✓. Type: Summable (fun x => ?r * C x) ✓ as I assumed. The isDefEq timeout on the hABC ascription — let me test WITHOUT the underscore: testC.mul_left 0.886. And test whether hAB.add alone with explicit coeffs works. Actually — I bet it's the `0.886` METAVAR in `testC.mul_left _` — the unify `?c * C x ≡ 0.886 * C x` should be instant... unless the unify goes higher-order and the elaborator descends into Real.log/Real.log computations?! No... Let me test variants in isolation quickly.
Reproduced in isolation. Testing which part of the chain is pathological.
tool: Bash
input
{
  "command": "cd D:/Python/Quantum/research/lean/NguyenRH1 && cat > /tmp/test_lam3.lean <<'EOF'\nimport Mathlib\nopen Complex Filter Topology MeasureTheory\nnoncomputable section\nnamespace TestNS\n\nlemma testA : Summable (fun \u2113 : \u2115 => (1:\u211d) / (((\u2113:\u211d)+14)+10)^2) := by sorry\nlemma testB : Summable (fun \u2113 : \u2115 => Real.log ((\u2113:\u211d)+14) / (((\u2113:\u211d)+14)+10)^2) := by sorry\nlemma testC : Summable (fun \u2113 : \u2115 => Real.log (Real.log ((\u2113:\u211d)+14)) / (((\u2113:\u211d)+14)+10)^2) := by sorry\n\n-- variant 1: explicit coefficient, no ascription on hAB\nlemma test3 : Summable (fun \u2113 : \u2115 =>\n      (0.433*0.1942+0.886*0.0736+2.9) * (1/(((\u2113:\u211d)+14)+10)^2)\n      + 0.433 * Real.log ((\u2113:\u211d)+14) / (((\u2113:\u211d)+14)+10)^2\n      + 0.886 * Real.log (Real.log ((\u2113:\u211d)+14)) / (((\u2113:\u211d)+14)+10)^2) := by\n  have hAB := (testA.mul_left _).add (testB.mul_left _)\n  exact hAB.add (testC.mul_left _)\n\nend TestNS\nEOF\nexport PATH=\"$HOME/.elan/bin:$PATH\" && time lake env lean /tmp/test_lam3.lean 2>&1 | head -8",
  "description": "Test variant without hAB ascription",
  "timeout": 300000
}
raw content_json (database row)
[{"type":"thinking","thinking":"REPRODUCED in isolation! The timeout is at 17:27 — `hAB.add (testC.mul_left _)` — so the .add itself times out. The hAB has an explicit ascribed type; hAB.add (testC.mul_left _) — the result type: Summable (fun x => (c0·A x + c1·B x) + c2·C x) vs hABC's ascription — the isDefEq times out. WHY?\n\nLet me think about `Summable.add`: `theorem Summable.add {f g : ι → M} (hf : Summable f) (hg : Summable g) : Summable fun x => f x + g x` — result: fun x => f x + g x with f,g as the implicit from the proofs. hAB : Summable (fun ℓ => c0·A ℓ + c1·B ℓ) — the ASCIBED type — so f := fun ℓ => c0·A ℓ + c1·B ℓ. testC.mul_left _ : Summable (fun x => ?c·C x) — with the underscore — the underscore ! The `testC.mul_left _` has a METAVARIABLE ?c — and when unifying hABC's ascription (0.886 coefficient), it should resolve... but maybe the elaborator tries to unify `fun x => ?c * C x` with the expected `fun x => 0.886 * C x` and the isDefEq on... — that should be trivial.\n\nOR: the result of hAB.add hC : Summable (fun x => hAB-fun x + hC-fun x) — with hAB-fun = the ascribed lambda and hC-fun = fun x => ?c·C x: the result's body: (c0·A x + c1·B x) + ?c·C x — vs ascription (c0·A x + c1·B x) + 0.886·C x — unify ?c := 0.886 ✓ — the isDefEq of the two big lambdas — the problem might be that hAB's ascribed TYPE is not reducibly equal to the body form... \n\nHmm wait — actually the deeper issue could be `testC.mul_left _`: Summable.mul_left's signature: `Summable.mul_left {f : ι → M} (r : M) (hf : Summable f) : Summable fun x => r * f x`? — or is r implicit/explicit? If `Summable.mul_left (r : M) (hf : Summable f)` — testC.mul_left _ — the FIRST explicit arg is r (a number, the underscore) and hf = testC ✓. Type: Summable (fun x => ?r * C x) ✓ as I assumed.\n\nThe isDefEq timeout on the hABC ascription — let me test WITHOUT the underscore: testC.mul_left 0.886. And test whether hAB.add alone with explicit coeffs works. Actually — I bet it's the `0.886` METAVAR in `testC.mul_left _` — the unify `?c * C x ≡ 0.886 * C x` should be instant... unless the unify goes higher-order and the elaborator descends into Real.log/Real.log computations?! No...\n\nLet me test variants in isolation quickly.","signature":""},{"type":"text","text":"Reproduced in isolation. Testing which part of the chain is pathological."},{"type":"tool_use","id":"Bash_206","name":"Bash","input":{"command":"cd D:/Python/Quantum/research/lean/NguyenRH1 && cat > /tmp/test_lam3.lean <<'EOF'\nimport Mathlib\nopen Complex Filter Topology MeasureTheory\nnoncomputable section\nnamespace TestNS\n\nlemma testA : Summable (fun ℓ : ℕ => (1:ℝ) / (((ℓ:ℝ)+14)+10)^2) := by sorry\nlemma testB : Summable (fun ℓ : ℕ => Real.log ((ℓ:ℝ)+14) / (((ℓ:ℝ)+14)+10)^2) := by sorry\nlemma testC : Summable (fun ℓ : ℕ => Real.log (Real.log ((ℓ:ℝ)+14)) / (((ℓ:ℝ)+14)+10)^2) := by sorry\n\n-- variant 1: explicit coefficient, no ascription on hAB\nlemma test3 : Summable (fun ℓ : ℕ =>\n      (0.433*0.1942+0.886*0.0736+2.9) * (1/(((ℓ:ℝ)+14)+10)^2)\n      + 0.433 * Real.log ((ℓ:ℝ)+14) / (((ℓ:ℝ)+14)+10)^2\n      + 0.886 * Real.log (Real.log ((ℓ:ℝ)+14)) / (((ℓ:ℝ)+14)+10)^2) := by\n  have hAB := (testA.mul_left _).add (testB.mul_left _)\n  exact hAB.add (testC.mul_left _)\n\nend TestNS\nEOF\nexport PATH=\"$HOME/.elan/bin:$PATH\" && time lake env lean /tmp/test_lam3.lean 2>&1 | head -8","description":"Test variant without hAB ascription","timeout":300000},"extra_content":null}]